Question

A 0.026 kg bullet is fired straight up at a falling wooden block that has a mass of 5.0 kg. The bullet has a speed of 750 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occured. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.

Answer 1

0.198 s

Step-by-step explanation:

Consider the motion of the block before collision

`v_(o)` = initial velocity of block as it is dropped = 0 m/s

`a` = acceleration = - g

`t` = time of travel

`v_(f)` = final velocity of block before collision

Using the kinematics equation

`v_(f) = v_(o) + at \\v_(f) = 0 + (-g)t\\v_(f) = - gt`

`m` = mass of the bullet = 0.026 kg

`v_(b)` = velocity of block just before collision = 750 m/s

`M` = mass of the block = 5 kg

`V` = final velocity of bullet block after collision = gt

Using conservation of momentum

`m v_(b) + Mv_(f) = (m + M) V\\(0.026) (750) + (5) (- gt) = (0.026 + 5) (gt)\\19.5 - 49 t = 49.2548 t\\t = 0.198 s`